#5) 3H2 + CO (g)->CH4 + H2O (g) ∆ H = ? for 300g CO
(1) 2H2(g) + O2(g) -> 2H2O(G) ∆ H = -483.6 kJ
(2) 2C(s) + O2(g) -> 2CO(g) ∆ H = -221.0 kJ
(3) CH4 +2O2(g) -> CO2(g) +2H2O(g) ∆ H = -802.7 kJ
(4) C(s) + O2(g) -> CO2(g) ∆ H = -393.5 kJ
(a) MULTIPLY (1) BY 3/2
(b) REVERSE (3)
(c) REVERSE AND MULTIPLY (2) BY ½
(d) STAYS THE SAME 3H2 + CO (g)->CH4 + H2O (g) + 205.7 kJ ∆ H = ? for 300g CO
Molar mass of carbon monoxide=12 + 1628g/mol
300g(1mol/28g)=10.7 mol
∆ H=(10.7 mol)(205.7 kJ/mol)=2200.99 kJ
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1.a. (i) C8H18 + 25/2 O2(g) -> 8CO2(g) + 9H2O (g) + 5.47 MJ
(ii) H2(g) + 1/2O2 -> H2O(l) + 285.8 kJ
(iii) C + O2 -> CO2 + 393.5 kJ/mol
b) 8C + 9H2 -> C8H18 (l)
(1) 8 TIMES (iii)
(2) 9 TIMES (ii)
(3) REVERSE (i)
∆ H = -250.2kJ
2) HCl(g) + NaNO2(s) -> HNO2(g) + NaCl(s) ∆ H=?
(i) 2NaCl + H2O -> 2HCl + Na2O ∆ H=507 kJ
(ii) NO + NO2 +Na2O -> 2NaNO2 ∆ H = -427 kJ
(iv) 2HNO2 -> N2O + O2 + H2O ∆ H = 34 kJ
(a) Reverse (i) and multiply by ½
(b) Reverse (ii) and multiply by ½
(c) Reverse (iv) and multiply by ½
(d) Multiply (iii) by ½ ∆ H = -78.5kJ
#4)
C2H2(g) + 2H2(g) -> C2H6(g) ∆ H for 200g ethyne?
1) C2H2(g) + 5/2O2 -> 2CO2(g) + H2O(l) ∆ H = -1299 kJ
2) H2 + ½ O2(g) --> H2O(l) ∆ H = -286 kJ
3) C2H6(g) + 7/2O2(g) -> 2CO2 (g) + 3H2O(l) ∆ H = -1560 kJ
a) 2 x (2)
b) Reverse (3)
c) (1) stays the same
a) 2H2 + O2(g) -> 2H2O(l) -572 kJ
b) 3H2O(l) + 2CO2(g) -> 7/2O2 + C2H6(g) +1560 kJ
c) C2H6(g) + 7/2O2(g) -> 2CO2 (g) + 3H2O(l) -1299 kJ
C2H2(g) + 2H2(g) -> C2H6(g) + 311 kJ
Molar mass of ethyne= 2(12) + 2 (1) = 26 g/mol
200g (1mol/26g) = 7.7mol
∆ H=(7.7mol)(311kJ/mol)=2392.3 kJ
#5)
3H2 + CO (g) ->CH4 + H2O (g) ∆ H = ? for 300g CO
(1) 2H2(g) + O2(g) -> 2H2O(G) ∆ H = -483.6 kJ
(2) 2C(s) + O2(g) -> 2CO(g) ∆ H = -221.0 kJ
(3) CH4 +2O2(g) -> CO2(g) +2H2O(g) ∆ H = -802.7 kJ
(4) C(s) + O2(g) -> CO2(g) ∆ H = -393.5 kJ
(a) MULTIPLY (1) BY 3/2
(b) REVERSE (3)
(c) REVERSE AND MULTIPLY (2) BY ½
(d) STAYS THE SAME
3H2 + CO (g) -> CH4 + H2O (g) + 205.7 kJ ∆ H = ? for 300g CO
Molar mass of carbon monoxide=12 + 1628g/mol
300g(1mol/28g)=10.7 mol
∆ H=(10.7 mol)(205.7 kJ/mol)=2200.99 kJ
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1.a. (i) C8H18 + 25/2 O2(g) -> 8CO2(g) + 9H2O (g) + 5.47 MJ
(ii) H2(g) + 1/2O2 -> H2O(l) + 285.8 kJ
(iii) C + O2 -> CO2 + 393.5 kJ/mol
b) 8C + 9H2 -> C8H18 (l)
(1) 8 TIMES (iii)
(2) 9 TIMES (ii)
(3) REVERSE (i)
∆ H = -250.2kJ
2) HCl(g) + NaNO2(s) -> HNO2(g) + NaCl(s) ∆ H=?
(i) 2NaCl + H2O -> 2HCl + Na2O ∆ H=507 kJ
(ii) NO + NO2 +Na2O -> 2NaNO2 ∆ H = -427 kJ
(iv) 2HNO2 -> N2O + O2 + H2O ∆ H = 34 kJ
(a) Reverse (i) and multiply by ½
(b) Reverse (ii) and multiply by ½
(c) Reverse (iv) and multiply by ½
(d) Multiply (iii) by ½
∆ H = -78.5kJ
#3) C2H5OH + O2 -> CH3COOH+ H2O
(i) Combustion of Ethanol: C2H5OH + 3O2 -> 2CO2 + 3H2O + 1367 kJ/mol
(ii) Combustion of acetic acid: CH3COOH + 2O2 -> 2CO2 + 2H2O + 875 kJ/mol
(a) C2H5OH + 3O2 -> 2CO2 + 3H2O ∆ H = - 1367 kJ/mol
(b) 2CO2 + 2H2O -> CH3COOH + 2O2 ∆ H = 875 kJ/mol
C2H5OH + O2 -> CH3COOH+ H2O + 492 kJ/mol ∆ H = - 492 kJ/mol
#4)
(1) HBr(aq) + KOH (aq) -> H2O(l) + KBr (aq) ∆ H = ? kJ
(2) KOH (s) -> KOH (aq) ∆ H = ? kJ
(3) KOH (s) + HBr (aq) -> H2O (l) + KBr(aq) ∆ H = ? kJ
a) Add (1) and (2) to get (3): HBr(aq) + KOH (aq) + KOH (s) -> H2O(l) + KBr (aq)+ KOH (aq)
b)
c)